Lemon

thanks.

i see i see.

okay apparently it is a definition first and foremost, so i guess we can do wtih it whatever we want.

@Thorgott its sometime stated as a theorem in proving its existence.

@Thorgott partition of unity.

@Thorgott because then the result is more general.

@Thorgott yeah i get that, but is that really it?

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In partition of unity, locally finite is assumed so that the final sum $\sum_i \theta_i(x) = 1$ is a finite sum . Why do we need this sum to be finite? Why is it a problem if the sum is not finite? Don't we only care about the sum equaling one?

ah ok.

I have a question about the definition of a local frame. A local frame for manifold $M$ of n-dim is an n- tuple that is linearly independent and spans the tangent bundle. Is there a reason why it isn't 2n-tuple? Since it needs to span the tangent bundle?

Ted. Would you be able to elaborate on your comment about the torsion = 0 being a condition on the connection and not on the bracket?

@TedShifrin I don't quite understand "the condition on the connection"

THe torsion operator is defined as $\tau(X,Y) = \nabla_X Y - \nabla_Y X - [X,Y].$ The bracket (Lie Derivative) is contructed independently of a connection. So why is it when the torsion is $0$, the bracket depends on the connection?

Should I not even bother thinking about expressions like $f(x_i) = (\sum f(x_i)x_i)(x_i)$? Because thinking about the free vector space as a space of function makes more symbollically sense than the polynoimal ring example.

it here being the space of formal sums?

You are right, I never actually questioned what is "addition" doing in the polynoimal rings. It just bothered me to see $f = \sum f(x_i)x_i$ and if i symbolically write $f(x_i) = (\sum f(x_i)x_i)(x_i)$ just did not make sense to me unless we really mean to write $\sum f(x_i) \delta_{x_i}$ since the free vector space is a space of functions.

@Jakobian I think your construction is a formal linear combination of a module (which has addition already). I am asking about formal linear combinations of any set, where "+_S" may not have any meaning. Most of the sources I am reading define it like this, but I just found one link that basically says if $ \sum f(x_i)x_i$ is not already value I want, then we force it to be $\sum f(x_i) \delta_{x_i}$

Why are formal linear combinations written as $f = \sum f(x_i)x_i$ instead of $f = \sum f(x_i) \delta_{x_i}$? The $x_i$s are elements of an arbitrary set $X$, where the space of formal linear combination is a set of maps.

Its all forms of $(0,\theta, \phi)$ since only the $d\theta^2 + d\phi^2$ part of the metric is what it describing $S^1$

Oh i see

@TedShifrin oh yes i forgot to add a missing cosine in my calculation. So the normal is just $\partial_\rho$. Normal to $\partial_x$ are just forms of $(0,y,z)$

i was doing some experiment. For $S^1$, the unit normal in Cartesian is $x^j/|x|\partial_j$. Converting this into spherical coordinates (is there a short cut other than just writing out the vector transformation for each $\partial_j$?), I got $N =\sin^2 \phi \partial_\rho$. Now I am thinking, with only one component, what tangent vector to the sphere is normal to this other than $0$?

oh right, one-to-one. Thanks.

It's a pair of an open set and a homemophism to R. But i think i m seeing what is wrong. If I try to define a chart on my $U_x$, I am trying to write a map $\phi(x = \pm \sqrt{1 - y^2},y) = y$. I think I am violating some part of the implicit function theorem and that's why my $(U_x, \phi)$ can't be a chart.

It would be $f(x,y) = \sqrt{1 - x^2 - y^2}$ where $(x,y) \in S^1$

Quick question, for $S^1$ the circle, why are the two sets $U_x = \{ (x,y) \in S^1: x \ neq 0 \}$ and $U_y = \{ (x,y) \in S^1: y\ neq 0 \}$ not charts but cutting them up into two pieces suddenly is? Are the sets $U_x $ and $U_y$ not homemoprhic to an open set in $R^n$?

Because the original definition of a smooth vector field $V$ along $\gamma$ is $V(t) \in T_\gamma M$. So when they say $N$ is a smooth vector field along $\gamma$, I thought they mean $N(t) \in T_\gamma M$

@TedShifrin I am not sure what in Calc 3 I need to revisit. I know the general definition of a vector field uses sections. But what i mean here is that $N$ would be section to the normal projection $\pi: NM \to M.$ The example says smooth vector field, which I assume means smooth tangent vector field.

@TedShifrin That's what I don't get. If it is normal, then it is not in the tangent space of $\gamma$, it is in a different space. So $N$ cannot be a smooth vector field along $\gamma.$

@TedShifrin sorry what is the calculus sense here? I am staring at $N$ and it is normal to the tangent vector field $(\gamma^{1'},\gamma^{2'})$, shouldn't it be in the normal space of $\gamma$?

IN fact since $N$ is normal to the tangent velocity vector, shouldn't this be in the normal space?

Question about an example. A vector field $V$ along a curve $\gamma$ is a continuous map such that $V(t) \in T_\gamma M.$ If $N(t) = R\gamma'(t)$ where $R$ is rotation by 90 degrees and $N(t) = (-\gamma^{2'},\gamma^{1'})$, then it is a smooth vector field along $\gamma.$ Why $N$ a vector field along $\gamma$? What part of it shows it is in the tangent space of $\gamma$?

No i m kinda learning this backwards from the abstract first. But some of this stuff I forget occasionally. Thanks for the pointer.

Oh i see I didn't know that. Actually I think that's how this stuff is introduced in the first place. I just kinda abandoned it along the way

@TedShifrin Sorry, isn't $\nabla$ always the connection on the ambient space? DId I write something that might imply my $\nabla$ might not be the connection on the ambient space? And the reason for my issue is, I've been looking at some second fundamental form calcs and practically everytime I do one of these, I only compute the normal part and the rest of the expansion i wrote out are completely pointless. I thought I was making a mistake.

Second Fundamental Form $A(X,Y) = g(\nabla_X Y, -\nu)$. Now $\nabla_X Y = (...)\partial_k$ where $(...)$ are the expansion components of the covariant derivative involve Christofell symbols and derivatives. If $g$ is just standard metric, wouldn't that mean the only surviving term ever is $g(\nu,\nu)$ where $\nu = \partial_n$ the last term?

If the answer is no, does the result change if $g$ is now the identity map?

If $f$ and $g$ are differentiable maps such that $| f(x) - g(y) | < \epsilon$ for any $\epsilon$, does this also imply there is a $\delta>0$ such that $|x - y| < \delta$

@Semiclassical Yes.

Guys if $g(x,y,z) = (u,v)$ and I want to find $\partial_{x} (f \circ g)$, it should be $\partial_x (f \circ g) = \partial_u f \partial_x u + \partial_v f \partial_x v$ right?

The covariant derivative $\nabla_X Y$ depends only on a the values of $Y$ on neighborhood of $p$ (and value of $X$ at $p$). Does this mean I can take covariant derivatives on $X$ and $Y$ that are only defined on an open set $U$ of the manifold $M$?

ah nvm, it is true. Because I can just break both vector fields down by its coordinate basis representation, say for $n=2$, $f(v^1 \partial_1 + v^2\partial_2)$ and $g(fv,w) = \sum fv^iw^i g_{ij}$

if $fv$ and $w$ are vector fields where $f$ is a smooth function, $g(fv,w) = fg(v,w)$ right?

@leslietownes oh ok. All the "examples" i keep thinking of are smooth functions. That's why it didn't make sense to me.

@Thorgott Might have step back a bit, why does taking the kth deriv yield a C^n -> C^{n-k} map?

@Thorgott Can you elaborate on the last point a bit about this map $C^n\rightarrow C^{n-k}$? But yes, now I also see the further consequences of what Ted wrote.

@Thorgott Yeah I was looking for the motivation behind it, but it turns out it is nothing more than iterating the construct on bounded and continuous functions space and that's why we are adding derivatives of f in the C^n norm and that's probably why I can't find many interesting analysis that talks about these C^n spaces because it is build from already known space, i.e. the C^0 space.

Does anyone know why the norm on C^n (space of continously differentiable functions) a sum of sups? |f|_C^n = sup|f| + sup|f'| + \dots + \sup|f^n|?